Question

You have a 175mg sample of Theobromine contaminated with 25mg (200mg total) of an impurity whose solubility in water at both 100 and 25C is half the solubility of Theobromine's. What is the maximum amount of milligrams of pure theobromine you will be able to recover from a recrystallization? What is your recrystallization yield in this case? Theobromine’s solubility in water is 6.6 mg/mL at 100 ºC and 0.5 mg/mL at 25 ºC.

Answer 1

At 100 degC, the minimum solvent, water is required to dissolve, 175 mg of Theobromine = 175/6.6 ml =26.51 ml of minimum water to be used.

Now, 3.3 mg of impurity would need 1 ml of water at 100 degC.

So, 25 mg of impurity would require = 25/3.3 ml =7.58 ml. Therefore the impurity would be completely dissolved at 100 degC.

Now, up on cooling to 25 degC, taking advantage of the solubility differences we can do the following.

To get rid of the impurity, minmum volume will be = 25/0.25 = 100 ml of water.

At 25 degC, the amount of Theobromine remained in the solution of 100 ml = 0.5*100 mg = 50 mg

So, the pure Theobromine recovered = (175 -50) mg = 125 mg.

So, the percentage recovery = 100*125/200 %= 62.5%